Numerical 4: Scaling

Perform magnitude and frequency scaling separately with w = 3 and ω _(C) = 300 and compute appropriate scaling for C_(new) as 10 µF

fig: Butterworth Filter

\[
\text{Given transfer function:} \quad T(s) = \frac{s + 0.5}{s + 3}
\]

\[
\text{General form:} \quad T(s) = \frac{s + \frac{1}{R_1 C_1}}{s + \left( \frac{1}{R_1 C_1} + \frac{1}{R_2 C_1} \right)} \tag{1}
\]

\[
\text{Comparing with:} \quad T(s) = \frac{s + 0.5}{s + 3} \tag{2}
\]

---

### Step 1: Compare Numerators
\[
\frac{1}{R_1 C_1} = 0.5 \quad \Rightarrow \quad R_1 C_1 = 2 \tag{3}
\]

---

### Step 2: Compare Denominators
\[
\frac{1}{R_1 C_1} + \frac{1}{R_2 C_1} = 3
\]

Substitute from equation (3):  
\[
\frac{1}{2} + \frac{1}{R_2 C_1} = 3 \quad \Rightarrow \quad \frac{1}{R_2 C_1} = 2.5
\]

---

### Step 3: Express \( R_1 \) in terms of \( R_2 \)

Multiply numerator and denominator:
\[
\frac{R_1 + R_2}{R_1 R_2 C_1} = 3
\]

Substitute \( R_1 C_1 = 2 \Rightarrow C_1 = \frac{2}{R_1} \) into the expression:

\[
\frac{R_1 + R_2}{2 R_2} = 3 \quad \Rightarrow \quad \frac{1}{2} + \frac{R_1}{2 R_2} = 3
\]

\[
\frac{R_1}{2 R_2} = 2.5 \quad \Rightarrow \quad R_1 = 5 R_2 \tag{4}
\]

---

### Step 4: Let \( C_1 = 1 \ \text{F} \), then from (3):
\[
R_1 = 2 \ \Omega
\]

Using equation (4):
\[
2 = 5 R_2 \quad \Rightarrow \quad R_2 = \frac{2}{5} = 0.4 \ \Omega
\]

---

###  Final Values:

\[
C_1 = 1 \ \text{F}, \quad R_1 = 2 \ \Omega, \quad R_2 = 0.4 \ \Omega
\]

fig: With Values

 

\text{The new circuit is given as:}

\[
C_{\text{new}} = 10 \ \mu\text{F}
\]

\[
C_{\text{new}} = \frac{C_{\text{old}}}{K_m}
\]

\[
K_m = \frac{C_{\text{old}}}{C_{\text{new}}} = \frac{1}{10 \ \mu\text{F}} = 10^5
\]

Given:  
\[
\omega = 3 \ \text{rad/sec}, \quad \omega(C) = 3000 \ \text{rad/sec}
\]

Frequency scaling factor:
\[
K_f = \frac{\omega(C)}{\omega} = \frac{3000}{3} = 1000
\]

---

\textbf{For Resistor:}  
\[
R_{\text{new}} = K_m \times R_{\text{old}}
\]

For \( R_1 = 2 \ \Omega \),
\[
R_{\text{new}} = 10^5 \times 2 = 2 \times 10^5 = 200 \text{ k}\Omega
\]

For \( R_2 = 0.4 \ \Omega \),
\[
R_{\text{new}} = 10^5 \times 0.4 = 4 \times 10^4 = 40 \text{ k}\Omega
\]

---

\textbf{For Capacitor:}  
\[
C_{\text{new}} = C_{\text{old}} \times \frac{1}{K_f \times K_m}
\]

Using given values,
\[
C_{\text{new}} = \frac{10 \ \mu\text{F}}{10^5 \times 10^3} = 10^{-3} = 1 \text{ mF}
\]

Or equivalently,
\[
C_{\text{new}} = \frac{1 \text{ F}}{10^3} = 10^{-3} = 1 \text{ mF}
\]

The Scaled circuit is given by

fig: Scaled filter